The flaw in the counter example is that they do not form a minimal generating set. We need prime order as we do not know that just because even if we do have a minimum generating set.
As to the proof - think...what would you normally do when something is true for (actually - easier than that - it's true for )? You will need the formula I gave though, as well as the two conditions I've given, above.
I have another question regarding such groups that I started to ponder but could not really get anywhere on...are all such groups soluble? If so, then we are done as either they are abelian and so we have that they are isomorphic to , or else the derived subgroup is a proper normal subggroup which we can quotient out and then apply the correspondence theorem (I believe we can anyway) to get the other subgroup.
However, I can't seem to prove that all such groups are soluble, but I can't find any counter-examples (admittedly my examples were far from many in number...emm, was all I could really think of). Clearly there are no non-abelian nilpotent examples, as their centres are trivial.
(a) H is a subgroup of G.
Then, by the hypothesis, as H is a group in its own right, . And
by applying the formula, I get and the proposition is proved. Unfortunately, I don't see how (a) and (b) can be proved.